When students learn multivariable calculus they're typically barraged with a collection of examples of the type "given surface X with boundary curve Y, evaluate the line integral of a vector field Y by evaluating the surface integral of the curl of the vector field over the surface X" or vice versa. The trouble is that the vector fields, curves and surfaces are pretty much arbitrary except for being chosen so that one or both of the integrals are computationally tractable.

One more interesting application of the classical Stokes theorem is that it allows one to interpret the curl of a vector field as a measure of swirling about an axis. But aside from that I'm unable to recall any other applications which are especially surprising, deep or interesting.

I would like use Stokes theorem show my multivariable calculus students something that they enjoyable. Any suggestions?

stokes theorem implies that the "angle form" on a sphere is not exact, [i.e. that the de rham cohomology of a sphere is non zero]. Thus corollaries include: brouwer fixed point, fundamental theorem of algebra, and absence of never zero vector fields on S^2. I gave all these applications in my first class on stokes theorem, since I myself had previously no idea what the theorem was good for. I was unable to deduce the jordan curve theorem, but perhaps that too is feasible, in the smooth case.
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roy smithMar 6 '13 at 19:36

8 Answers
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If you don't mind specializing Stokes theorem to Green's theorem, then one of the most practical applications is computation of the area of a region by integrating around its contour.
I am old enough to have used a planimeter, a delightful physical embodiment of Green's theorem:
One can also derive an (otherwise non-obvious) formula for the area of a planar polygon
via Green's theorem: $A = \frac{1}{2} \sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i$.

In the theory of electromagnetism, the classical Stokes Theorem moves between the differential and integral forms of two of Maxwell's four equations; see https://en.wikipedia.org/wiki/Stokes%27_theorem#In_electromagnetism for discussion. Note that the integral forms may be directly interpreted using classical physical intuition, while the differential forms give us differential equations that we might solve, so it is important that we can switch between them.

ETA: I think that Wikipedia's discussion is a little vague, although possibly appropriate in that context. So here is more detail, looking at Faraday's Law. In terms of physically observable quantities, the law states that the rate of change of the magnetic flux through a stationary surface is proportional to the electromotive force around the boundary of the surface. The magnetic flux is the surface integral of the magnetic field $ \vec H $, and the EMF is the line integral of the electric field $ \vec E $, so we have $$ \oint _ { \partial S } \vec E \cdot \mathrm d \vec r = - \frac { \mathrm d } { \mathrm d t } \iint _ S \vec H \cdot \mathrm d ^ 2 \vec A $$ using standard units and sign conventions. Applying the classical Stokes Theorem on the left and using that $ S $ is stationary on the right, this becomes $$ \iint _ S ( \nabla \times \vec E ) \cdot \mathrm d ^ 2 \vec A = - \iint _ S \frac { \partial \vec H } { \partial t } \cdot \mathrm d ^ 2 \vec A \text ; $$ since this holds for arbitrarily small surfaces, we conclude that $$ \nabla \times \vec E = - \frac { \partial \vec H } { \partial t } \text , $$ a differential equation. (The argument in reverse is even easier, since you don't have to worry about arbitrarily small surfaces.)

Sure, I even consider that to be the main application. But like Joseph O'Rourke's example this uses Stokes theorem restricted to closed curves and regions and vector fields on the plane. Also, hard to fit into a multivariable calculus class.
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Jonah SinickDec 20 '11 at 23:42

A student may also learn about the content from Stokes theorem from instances where it failed to hold as expected. For example, one has to exercise care when trying to use the theorem on domains with holes. Turn this around: the failure of Stokes to hold as expected tells you about the cohomology of the domain. I think it is possible via concrete examples to illustrate this point in a multivariate calculus class without using the more technical phraseology.

There's a nice book called From calculus to cohomology by Ib Madsen and Jørgen Tornehave with exactly this viewpoint. This can possibly be a used as a supplement for a strong and interested audience.
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Daniel LarssonDec 21 '11 at 8:16

Daniel, thank you for the reference!
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Nilima NigamDec 21 '11 at 16:10

You can tell your students that a clever use of Stokes theorem can give you a Fields medal. Indeed, the proof that the formality map given by M. Kontsevich is a $L_\infty$-morphism, is nothing else than Stokes theorem. A detailed account of this can be found in Deformation quantization of Poisson manifolds. Lett. Math. Phys. 66 (2003), no. 3, 157–216 or with more details in Déformation, quantification, théorie de Lie, 123–164, Panor. Synthèses, 20, Soc. Math. France, Paris, 2005 which is in English, contrary to its title.

A nice application in fluid mechanics is Kelvin's circulation theorem. You could also discuss how it fails to hold, if there are obstacles in the fluid flow. In the same spirit stokes theorem is applied in the canonical formalism of classical mechanics to find the poincare-cartan integral invariants.

I find interesting that divergence theorem (that is a corollary of the general Stokes theorem on differentiable varieties) in a vectorial form (one integral for any cathesian cohordinate) give a proof of the Archimedes' principle of buoyancy on the fluid .

given a body immersed on a (incompressible) fluid let $S$ its surface and for each point $p\in S$ let $\overrightarrow{n}:=(n_x(p), n_y(p), n_z(p))$ the normal versor to $S$ in $p$. then the total force on the body is the (vector) integral $\int_S\mu\cdot (L-z(p))\cdot -\overrightarrow{n}\cdot dS=$

then from the divergence theorem (on each of the three components) and from $\nabla\circ((l-z)\overrightarrow{i})=\partial/\partial x (l-z)=0,\ \nabla\circ((l-z)\overrightarrow{j})=\partial/\partial y (l-z)=0$
$\nabla\circ((l-z)\overrightarrow{k})=\partial/\partial z (l-z)=-1$